# NEET Chemistry The Solid State Questions Solved

A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ............ Any packing of spheres leaves out voids in the lattice. The percentage by volume of empty space of this is

(1) 26%

(2) 21%

(3) 18%

(4) 16 %

(1) The hexagonal base consists of six equilateral triangles, each with side 2r and altitude 2r sin 60°.

Hence, area of base = $\text{6\hspace{0.17em}}\left(\frac{1}{2}\left(2r\right) \left(2rsin{60}^{o}\right)\right)=6\sqrt{3}.{r}^{2}$

The height of the hexagonal is twice the distance between closest packed layers.

The latter can be determined to a face centred cubic lattice with unit cell length a. In such a lattice, the distance between closest packed layers is one third of the body diagonal, i.e. $\frac{\sqrt{3}a}{3}$

Hence

Now, in the face centred lattice, atoms touch one another along the face diagonal,

Thus, $4r\text{\hspace{0.17em}}=\sqrt{2}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}a$

With this, the height of hexagonal becomes :

Volume of hexagonal unit is, V = (base area) × (height) $=\text{\hspace{0.17em}}\left(6\sqrt{3}\text{\hspace{0.17em}}{r}^{2}\right)\text{\hspace{0.17em}}\left[\frac{4\sqrt{2}}{\sqrt{3}}.\text{\hspace{0.17em}}r\right]=24\sqrt{2}.\text{\hspace{0.17em}}{r}^{3}$

In one hexagonal unit cell, there are 6 atoms as described below :

• 3 atoms in the central layer which exclusively belong to the unit cell.

• 1 atom from the centre of the base. There are two atoms of this type and each is shared between two hexagonal unit cells.

• 2 atoms from the corners. There are 12 such atoms and each is shared amongst six hexagonal unit cells.

Now, the volume occupied by atoms = $6\left[\frac{4}{3}\pi {r}^{3}\right]$

Fraction of volume occupied by atoms $=\frac{6\left(\frac{4}{3}\pi {r}^{3}\right)}{24\sqrt{2}.\text{\hspace{0.17em}}{r}^{3}}=\pi /3\sqrt{2}=0.74.$

Fraction of empty space = $\left(1.00-0.74\right)=0.26$

Percentage of empty space = 26%

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